# Show $\lbrace\phi(X_{n})\rbrace$ is a submartingale if $\lbrace X_{n}\rbrace$ is

If $\lbrace X_{n}\rbrace$ is a submartingale with respect to filtration $\lbrace \mathcal{F}_{n}\rbrace$, and $\phi:\mathbb{R}\to\mathbb{R}$ is convex and increasing, then $\lbrace\phi(X_{n})\rbrace$ is a submartingale, too.

1. By convexity of $\phi$ and the fact that $X_{n}$ is $\mathcal{F}_{n}$-measurable, $\phi(X_{n})$ is $\mathcal{F}_{n}$-measurable, thus the new process is adapted to $\lbrace \mathcal{F}_{n}\rbrace$.
2. To verify that $\mathbb{E}(\phi(X_{n+1})|\mathcal{F}_{n})\geq\phi(X_{n})$, Jensen's inequality can be applied.
3. Now my question: How to show that $\mathbb{E}(\phi(X_{n})^{+})<\infty$?
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 You need integrability as another condition. Otherwise consider some random variable with zero mean, but infinite third moment, and $\phi(x) = x^2 1_{x \geq 0}$. – Thomas Feb 4 at 13:45